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Problem set — the second law and heat engines

Problem set: the second law and heat engines

Exam-style problems. Work them with a pen before touching a computer; the last one is meant to be finished numerically. Solutions and marking rubrics live with the instructor material and are deliberately not on this site.

Take kB=1.381×10−23 J K−1\kB = 1.381 \times 10^{-23}\ \mathrm{J\,K^{-1}} throughout, and give every temperature in kelvin before doing anything else with it.

Problem 1 — the two statements

(a) State the Kelvin and Clausius forms of the second law, and explain what the words cyclic and sole result are doing in each. Give an example of a real process that converts heat entirely into work, and say why it does not violate the Kelvin statement.

(b) A device is claimed to draw 500 J500\ \mathrm{J} per cycle from a single reservoir at 400 K400\ \mathrm{K} and deliver 500 J500\ \mathrm{J} of work, returning to its initial state each cycle. Construct, in words and with an energy diagram, a machine that combines this device with an ordinary refrigerator to violate the Clausius statement.

(c) Explain why a gas undergoing a single quasistatic isothermal expansion is not a counterexample to (a), even though ΔU=0\Delta U = 0 and all the heat absorbed becomes work.

Problem 2 — bounds on a real plant

A steam turbine takes heat from a boiler at 811 K811\ \mathrm{K} and rejects it to a condenser at 311 K311\ \mathrm{K}. It absorbs 2.40 GJ2.40\ \mathrm{GJ} of heat per second.

(a) Compute the maximum possible efficiency and the maximum possible power output.

(b) The plant actually delivers 0.95 GW0.95\ \mathrm{GW}. Compute its efficiency, and the rate at which it dumps heat into the condenser.

(c) An engineer proposes raising the output by cooling the condenser to 280 K280\ \mathrm{K} with a refrigeration unit, which would itself reject its heat to the surroundings at 311 K311\ \mathrm{K}. Without computing anything, give the argument for why this cannot pay for itself. Then estimate the minimum work the refrigeration unit would need — assume it must move the full condenser load — and compare that with the extra output the colder condenser would buy.

(d) Compute 1−Tc/Th1 - \sqrt{T_c/T_h} for this plant and compare it with your answer to (b). Comment on what this does and does not tell you.

Problem 3 — Carnot’s theorem as a tool

(a) An inventor claims an engine running between 500 K500\ \mathrm{K} and 300 K300\ \mathrm{K} that absorbs 1000 J1000\ \mathrm{J} and delivers 450 J450\ \mathrm{J} per cycle. Is the claim possible? Show your reasoning, and state which law is violated if it is not.

(b) A second inventor claims 380 J380\ \mathrm{J} from the same 1000 J1000\ \mathrm{J} between the same reservoirs. Is this possible? Is it necessarily achievable?

(c) Two reversible engines run between the same reservoirs, one using helium and one using steam. Prove, using only Carnot’s theorem and without evaluating anything, that they have identical efficiencies. Where exactly would a proof that depended on the working substance break down?

(d) Explain how a reversible engine could be used to construct a temperature scale, and state one practical reason this is not how thermometers are actually built.

Problem 4 — the refrigerator and the kitchen

A refrigerator maintains its interior at 275 K275\ \mathrm{K} in a kitchen at 298 K298\ \mathrm{K}.

(a) Compute the maximum possible coefficient of performance. The unit actually achieves 3.2; compute the electrical power it needs to remove heat at 150 W150\ \mathrm{W}.

(b) Compute the rate at which the refrigerator delivers heat into the kitchen. A student suggests cooling the kitchen on a hot day by leaving the refrigerator door open. Say exactly what happens to the kitchen’s temperature, and why.

(c) The same hardware is used as a heat pump to warm a house at 295 K295\ \mathrm{K} from outside air at 273 K273\ \mathrm{K}. Compute the maximum COP. Compare the electricity needed to deliver 5 kW5\ \mathrm{kW} of heating this way with the electricity a resistive heater would need, and explain in one sentence where the difference comes from.

(d) The outside temperature drops to 258 K258\ \mathrm{K}. Recompute the maximum COP and comment on why heat pumps are harder to justify in very cold climates.

Problem 5 — where the work went

An engine runs between reservoirs at 600 K600\ \mathrm{K} and 300 K300\ \mathrm{K}, but its heat exchangers are imperfect: the working gas sits at 560 K560\ \mathrm{K} while absorbing heat and at 340 K340\ \mathrm{K} while rejecting it. Internally every stroke is reversible.

(a) Compute the engine’s efficiency and compare it with the Carnot bound for the reservoirs.

(b) Per 1000 J1000\ \mathrm{J} absorbed from the hot reservoir, compute the work delivered, the heat rejected, and the value of ∮δQ/Tres\oint \dbar Q / T_{\text{res}}, being careful about which temperature belongs in each denominator.

(c) Show that the work lost, compared with a perfect Carnot engine drawing the same 1000 J1000\ \mathrm{J}, equals TcT_c times the quantity you computed in (b). Interpret this result.

(d) Explain why closing the temperature gaps to zero is not a design improvement anyone can adopt, and what is actually being traded away.

Problem 6 — computational

Using the module’s laboratory notebook and the thermolab.engines module:

(a) Build Carnot cycles between 600 K600\ \mathrm{K} and 300 K300\ \mathrm{K} for at least four combinations of working gas (f=3,5,6f = 3, 5, 6) and expansion ratio, and tabulate the efficiencies. Report how many figures they agree to, and say what that is evidence for.

(b) Using endoreversible_cycle, plot the efficiency against the temperature gap for gaps from 0 to 100 K100\ \mathrm{K}, with the Carnot bound drawn as a horizontal line. At what gap has the engine lost a quarter of its available efficiency?

(c) On the same axes plot the entropy produced per cycle against the gap. Fit its behaviour at small gap and state the power law you find.

(d) Attempt to construct any cycle from the library that exceeds 1−Tc/Th1 - T_c/T_h. Report what you tried. Then state precisely why failing to find one is not a proof of Carnot’s theorem, and what it is evidence of instead.

Problem 7 — the cost of irreversibility

(a) A block sliding across a table at 300 K300\ \mathrm{K} is brought to rest by friction, converting 50 J50\ \mathrm{J} of kinetic energy into heat that ends up in the table. Show that undoing this process without trace would violate the Kelvin statement, and compute the entropy produced.

(b) On a winter day, 500 J500\ \mathrm{J} of heat leaks through a window from a room at 293 K293\ \mathrm{K} to the outside air at 263 K263\ \mathrm{K}. Compute the entropy produced, and the work that an ideal engine placed across the window could have delivered from the same heat. Show that the two answers are related as the lost-work theorem says they must be.

(c) One mole of ideal gas (N=6.022×1023N = 6.022 \times 10^{23}) at 300 K300\ \mathrm{K} expands freely into a vacuum, doubling its volume inside an insulated box. Compute its entropy change by choosing a suitable reversible process between the same end states, and explain why that is legitimate although the real process is nothing like it. How much work would the reversible process have delivered, and how is that number related to the entropy produced?

(d) Two identical blocks, each of heat capacity C=500 J K−1C = 500\ \mathrm{J\,K^{-1}}, one at 400 K400\ \mathrm{K} and the other at 200 K200\ \mathrm{K}, are put in thermal contact inside an insulated box. Find the final temperature and the total entropy change. Then show, for any starting temperatures T1T_1 and T2T_2, that

ΔS=Cln⁡(T1+T2)24 T1T2≥0,\Delta S = C \ln \frac{(T_1 + T_2)^2}{4\, T_1 T_2} \ge 0 ,

with equality only when T1=T2T_1 = T_2.

(e) Suppose instead that a reversible engine runs between the two blocks of (d) until they reach a common temperature. Find that temperature and the total work delivered, and explain in one sentence why the final temperature is lower than in (d).

Problem 8 — the Otto engine

An idealised petrol engine takes in air at T1=300 KT_1 = 300\ \mathrm{K} and compresses it adiabatically by a factor r=10r = 10; treat air as an ideal gas with γ=7/5\gamma = 7/5. Burning the fuel then heats it at constant volume to T3=2200 KT_3 = 2200\ \mathrm{K}, it expands adiabatically back to its original volume, and it gives up its remaining heat at constant volume.

(a) Show that the efficiency is η=1−r1−γ\eta = 1 - r^{1-\gamma}, independent of T3T_3, and evaluate it.

(b) Find the temperatures T2T_2 after compression and T4T_4 after expansion. Compare the efficiency with the Carnot bound between the coldest and hottest temperatures the air reaches.

(c) Every stroke of this cycle is reversible, yet it falls short of that bound. Explain why, by dividing the cycle into thin slices with adiabats.

(d) Compute the entropy change of the air on the heating stroke and on the cooling stroke. Dividing each stroke’s heat by its entropy change defines an average temperature for that stroke; compute both, and show that the engine’s efficiency is that of a Carnot engine between them.

(e) The efficiency grows with rr. What stops petrol engines from using much higher compression ratios, and how does the diesel engine of module 06 get around it?